Re: Basic AC wattage question: am I doing my math right?

On Sat, 29 Dec 2007 00:08:08 -0800 (PST), HC <hboothe@xxxxxxx> wrote:




Your math is right for a simple (ie, resistive) load, but might be off
by, say, 2:1 in either direction, depending on the actual load
waveform and/or power factor.

John

Hey, John, the more I've been reading the more I think my math and
measurements have been badly wrong because they're rooted in DC
methodology which seems to be quite different from the AC world (or
can be for equipment/circuits that are not 100% resistive). So, as
you say, I could be quite wrong by just simply not knowing the
intricacies of AC power flow. I found a formula for calculating AC
current consumption on single-phase which is P = V x I x cosine Theta
which is great except that Theta is the "power factor angle" of the
equipment which I don't currently know. I approached this problem
thinking like I do about DC and I've found that it is wrong and the
problem is considerably more complex. I'm going to do a lot more
reading about this and "reboot" my whole test.

Thank you for your reply.


I checked my 20 inch Sony and it was drawing 2 watts when turned off, but 5 VA
(volt-amperes, the product of separately measured volts and amps; the same thing
you did).

The power was measured with a Yokogawa electromechanical wattmeter specifically
designed to measure power when the power factor is low.

These two measurements mean that the power factor was about .4.

What you should know is that the measurement you made is an upper bound for the
true power. For example, if the power factor (which you don't know) is .33333,
then the true power consumption is 1/3 of what you measured and you are paying
less than 15 cents a month for that TV's idle power; more like 5 cents. Do you
really need to know more than that? If so, then get a Kill-a-watt, but be aware
that the Kill-a-watt will have inadequate resolution for really accurate
measurements at such low powers.
.

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